enclosed room can be estimated by using other
approximations. If we assume that the whole inner surface of
a room (e.g. floor area of 30 m2, height of 2.5 m) is irradiated
with the emission of an infrared source with an overall
irradiance of Ee = 0.4 mW/m2, then an emitted radiant flux of
50 mW is necessary (surface = 115 m2, 100 % efficiency).
With an 80 % reflection loss, about 250 mW of emitted
radiation will be required for reliable reception in the whole
room.
250 mW is a value, which can be achieved with an emitter
TSAL6400 operating at a peak forward current of 700 mA.
Under these conditions, no direct path between the emitter
and the receiver is assumed, but that radiation will reach the
detector after at least one reflection. This kind of remote
control system is very user friendly for the customer because
he can aim the handset in any direction of his living room. An
IR emitter with a wide emitting angle will also provide this
kind of comfortable remote control system.
Vodio sam se ovim uputama koje sam nasao nasao u nekim Vishay dokumentima, ali imam problem gde nemogu nikako da postignem da proguram 700mA kroz diodu. Znam da duty cycle mora da mi bude jako mali da ne bi sprzio diodu, ali to za sada nije problem, zato sto ne mogu ni da dobijem tih 700mA nikako.
Koristim IR LED TSAL6400, i 2n2222 tranzistor, 5 v napajanje i otpornik(7.5 i 10oma).
Sema izgleda ovako:
5V ide na otpornik i IR diodu, i onda to ide u collector tranzistora, emitter mi je povezan na nulu. A sad na bazu mu saljem 5V square wave iz PIC-a. Maksimum sto mogu da dobijem kroz IR diodu je 150-180mA sa otpornikom od 10 do 7.5 oma.
Pad napona izmedju collectora i emittera je 2.2v, 1.3v je pad napona na diodi, sto ostavlja 1.5v na otporniku. Tim sam dosao do cifre of 150mA koja prolazi kroz diodu. Kako da dobijem tih 700mA koji su mi potrebni da upalim diodu?